Re: Skolem's Paradox and why is math the way it is?

From: J.E. (troubled6man_at_yahoo.com)
Date: 10/09/04 

Date: 9 Oct 2004 07:14:09 -0700

daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<ck6s1p0v1b@drn.newsguy.com>...
> J.E. says...
>
> >What is the cardinality of the smallest sets that cannot be named?
>
> There are countable sets that cannot be named (using ZF).

So there are real numbers that can't be named, uncountably many in
fact because only countably many are named, right?

> >I don't that there is REALLY an existance proof for sets that cannot be
> >proved to exist.
>
> I don't know what that means. As I said,
>
> ZF proves there exists a set that is uncountable.

ZF proves the LACK of a bijection IN the ZF system, yes. That doesn't
mean that one couldn't add introduce a new bijection and a new real
that weren't there previously, in fact one CAN.
 
> >I think you are using language in a sloppy manner.
>
> That's the reason people use mathematics. The claim that the
> set of reals is uncountable is perfectly precisely rendered into
> mathematics, and is provably true.

The lack of a bijection is provably true, but that doesn't mean it is
was impossible, it means that it wasn't derived from the axioms.
 
> >Proving a lack of a bijection proves the class of bijections is small,
> >NOT that the class of subsets is large.
>
> Like I said, to say that a set is uncountable is to say that
> there is no bijection between that set and N.

If you look at the axiom of equality two sets are the same IFF they
have the same elements. There exist subsets that one can neither
prove are inthe power set, not prove that they are (because one can't
prove they exist as sets as all). Let R- be the set of reals that
does NOT contain any subset of naturals that cannot be proven to exist
in ZF, and let R+ be the set that DOES contain them all. R- is pure,
but R+ is not. But sadly, ZF cannot distinguish between R- and R+
because the axiom of equlaity is OUT TO LUNCH. R- COULD be countable
since all you need to do is add the existance of a bijection B from N
to R- to the ZF axioms (which can consistently be done, let's called
the new axioms ZF'). R- is not the set of all subsets of N that can
be proven to exist in ZF', that would be a new set (that you can't
prove existed in ZF) called R'-, but its definition as the set of all
subsets that can be proven to exist in ZF still stands.

Do you understand that R+ may or may NOT be truly uncountable, but
that R- is only uncountable because one OMITTED a consistent countable
set that COULD have consiostently been in the ZF axiom system. What I
object to is that people people claim that R is REALLY the set R+
because "all" subsets exist, but then at the same time they deny the
existance of the set B. You can't have it both ways!

> >Look at the countable model to see this result.
> >This REALLY isn't "falling off a log" kind of
> >obvious to you?
>
> A countable model of ZF obviously doesn't contain all the reals,
> because the set of reals is uncountable.

You haven't distinguished between R- and R+. And the axiom of
equality doesn't distinguish either. This is a serious prroblem if
you are going to say that elements of R+ are the "real" reals. What
is your basis for that claim?

Relevant Pages

  • Re: Calculus XOR Probability
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  • Re: Calculus XOR Probability
    ... axiom system has to be accepted until it is proved, ... The invertible functions of IFR map reals to reals. ... TO means they are a bijection between the subset of the reals ... functions between the naturals and some other Peano set. ...
    (sci.math)
  • Re: Zenkins paper on Cantor
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  • Re: Zenkins paper on Cantor
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    (sci.math)
  • Re: Skolems Paradox and why is math the way it is?
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    (sci.math)